Animated Solution for Mathematics - Straight Lines: The vertices of a triangle are $[a{t}_1{t}_2,a(t_1+t_2)]$, $[a{t}_2{t}_3,a(t_2+t_3)]$, $[a{t}_3{t}_1,a(t_3+t_1)]$. Find the orthocentre of the triangle.

Welcome, fellow explorer of the mathematical universe! Today, we are going to unravel a problem that looks like a dense thicket of algebra but is actually a beautiful, elegant geometric dance.

We are tasked with finding the orthocenter of a triangle whose vertices are defined by the parameters $t_1,t_2,$ and $t_3$. The vertices are given by: $A=(a{t}_1{t}_2,a(t_1+t_2))$ $B=(a{t}_2{t}_3,a(t_2+t_3))$ $C=(a{t}_3{t}_1,a(t_3+t_1))$

These are not random points; they are the intersection points of tangents to the parabola $y^2=4ax$. By recognizing this, we have already won half the battle.

Our mission is to find the orthocenter, $H$, which is the point where the altitudes of our triangle meet. To find this point, we need the equations of at least two altitudes.

Let's start with the altitude from vertex $A$. To find its equation, we first need the slope of the opposite side, $BC$. Using the slope formula $m=\frac{y_2-y_1}{x_2-x_1}$, we calculate:

Here is where the magic happens. If we factor out $a$ from the numerator and $a{t}_3$ from the denominator, we get:

The term $(t_2-t_1)$ cancels out beautifully, leaving us with $m_{BC}=\frac{1}{t_3}$.

Now, we know that the altitude from $A$ is perpendicular to $BC$. The condition for perpendicularity is that the product of their slopes must be $-1$. Therefore, the slope of our altitude, $m_{altA}$, must be $-t_3$.

With this slope and the coordinates of vertex $A$, we use the point-slope form: $y-a(t_1+t_2)=-t_3(x-a{t}_1{t}_2)$

Expanding this, we get $y-a(t_1+t_2)=-x{t}_3+a{t}_1{t}_2{t}_3$. Rearranging to a standard form, we arrive at our first equation:

Now, we could repeat this entire process for the altitude from $B$, but let's be clever. By observing the symmetry of our coordinates, we can immediately write the equation for the second altitude.

The slope of $AC$ is $1/t_1$, so the altitude from $B$ has a slope of $-t_1$. Following the same logic, the equation for the altitude from $B$ is:

Now we have a system of two linear equations. Subtracting the second from the first, the $y$ terms and the complex cubic terms vanish, leaving us with $x(t_3-t_1)=a(t_1-t_3)$.

Dividing by $(t_3-t_1)$, we find the stunningly simple result: $x=-a$.

We have found the $x$-coordinate of the orthocenter! Substituting $x=-a$ back into our first equation, we solve for $y$:

This simplifies to $y=a(t_1+t_2+t_3)+a{t}_1{t}_2{t}_3$. Thus, the orthocenter $H$ is:

But look at that $x$-coordinate again. It is $-a$. In the world of parabolas, $x=-a$ is the directrix.

We have just proven that the orthocenter of a triangle formed by three tangents to a parabola always lies on its directrix. Isn't that magnificent? Keep this insight in your toolkit; it is the kind of deep, structural understanding that separates the good from the great in JEE Advanced.